$\overline{AC}$ is $6$ units long $\overline{BC}$ is $3$ units long $\overline{AB}$ is $3\sqrt{5}$ units long What is $\cos(\angle BAC)$ ? $A$ $C$ $B$ $6$ $3$ $3\sqrt{5}$
Solution: SOH CAH TOA os = djacent over ypotenuse adjacent $= \overline{AC} = 6$ hypotenuse $= \overline{AB} = 3\sqrt{5}$ $\cos(\angle BAC )=\frac{6}{3\sqrt{5}}$ $=\dfrac{2\sqrt{5} }{5}$